Question

(a) Draw a ray diagram to show the image formation by a combination of two thin convex lenses in contact. Obtain the expression for the power of this combination in terms of the focal lengths of the lenses.
(b) A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when the angle of incidence is $\frac{3}{4}$ of the angle of prism. Calculate the speed of light in the prism.

Answer 1

For first lens ,
$\frac{1}{f_1}=\frac{1}{v^{\prime }}-\frac{1}{u}\left(i\right)$
For second lens ,
$\frac{1}{f_2}=\frac{1}{v}-\frac{1}{v^{\prime }}\left(ii\right)$
Adding (i) and (ii),
$\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{v}-\frac{1}{u}\left(iii\right)$
If f is the combined focal length then
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
From equation (iii),
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$
$P=\frac{1}{f_1}+\frac{1}{f_2}[\because \frac{1}{f}=P$]
$P=\frac{f_1+f_2}{f_1{f}_2}$
(b) Given : A= ${60}^{\circ }$
i = $\frac{3}{4}A$
$i=\frac{3}{4}\times {60}^{\circ }$
$i={45}^{\circ }$
For angle of minium deviation,
$r=A/2=\frac{{60}^{\circ }}{2}r={30}^{\circ }\because \mu =\frac{sini}{sinr}$
$\frac{c}{v}=\frac{sin{45}^{\circ }}{sin{30}^{\circ }}[\because \mu =\frac{c}{v}]$
$\frac{3\times {10}^8}{v}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\frac{3\times {10}^8}{v}=\sqrt{2}v=\frac{3\times {10}^8}{1.414}m/sv=2.12\times {10}^{8}m/s$